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2 years ago

Halppppppp meeeaaaaaaaaaaaa

Mathematics

1 answer:

zvonat [6]2 years ago

7 0

Answer:

The x-intercept is the location on the graph when the output is 0.

f(x) > 0 is intervals of the domain where the graph is above the x-axis.

y-intercept is the location on the graph when the input is 0.

f(x) < 0 is intervals of the domain where the graph is below the x-axis.

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Evaluate the expression for a = 11, b = 28. c = 25, d = 10: a + b - C-d

Soloha48 [4]

Answer:

Step-by-step explanation:

11 + 28 - 25 - 10

39 - 25 - 10

14 - 10

Remember PEMDAS (Parentheses, Exponents, Multiply, Divide, Add, Subtract)

4 0

3 years ago

What the the perimeter of this polygon?

DENIUS [597]

Answer:

21.7

Step-by-step explanation:

you add all along the shape and it adds up to 21 and 21.7 is the only answer that has 21 in it

hope this helps.

6 0

2 years ago

Estimate the time it takes for a free fall drop from 10 meters height. Also estimate the time a 10 m platform diver would be in

murzikaleks [220]

Answer:

a) $t = \sqrt{\frac{2 *10m}{9.8 m/s^2}}=1.43s$

b) $4.9t^2 -2t -10 =0$

And we can use the quadratic formula to solve it:

$t = \frac{2 \pm \sqrt{(-2)^2 -4(4.9)(-10)}}{2*4.9}$

$t =\frac{2 \pm \sqrt{200}}{9.8}$

$t_1 =1.65 s, t_2 =-1.24 s$

And since the time can't be negative the correct option would be $t=1.65 s$

Step-by-step explanation:

For this case we can use the following kinematics formulas:

$y_f = y_o + V_o t + \frac{1}{2} a t^2$

For this case we assume that the only acceleration is the gravity a = g =9.8 m/s^2. And for this case we can assume that the reference point is $y_o =0$ and the final height would be $y_f = -10 m$ since is below the initial point.

Teh acceleration would be a=-g, since the gravity is acting dowward, we assume that the initial velocity is 0, so then we have everything to replace and we got:

$-10 m = 0m + (0m/s)t -\frac{1}{2} (9.8 m/s^2) t^2$

And solving for t we got:

$t = \sqrt{\frac{2 *10m}{9.8 m/s^2}}=1.43s$

For the second part assuming that We have an initial vertical speed of $v_o = 2 m/s$ we have the following equation:

$-10 m = 0m + (2m/s)t -\frac{1}{2} (9.8 m/s^2) t^2$

And we have this quadratic equation:

$-10 = 2t -4.9t^2$

$4.9t^2 -2t -10 =0$

And we can use the quadratic formula to solve it:

$t = \frac{2 \pm \sqrt{(-2)^2 -4(4.9)(-10)}}{2*4.9}$

$t =\frac{2 \pm \sqrt{200}}{9.8}$

$t_1 =1.65 s, t_2 =-1.24 s$

And since the time can't be negative the correct option would be $t=1.65 s$

6 0

3 years ago

What is the approximate area of the circle shown (please help first answer gets brainliest)

Allisa [31]

Answer:

The answer is B.

Step-by-step explanation:

8 0

2 years ago

Find the two-digit number satisfying the following two conditions:

lina2011 [118]

--------------------------------------
Define x :
--------------------------------------
Let the two digit number be 10x + y.

--------------------------------------
Construct Equation :
--------------------------------------
Four times the units digit is six less than twice the tens digit
⇒ 4y = 2x - 6
⇒2y = x - 3
⇒ x = 2y + 3

The number is nine less than three times the number obtained by reversing the digits.
⇒10x + y = 3(10y + x) - 9
⇒ 10x + y = 30y + 3x - 9
⇒ 7x = 29y - 9

--------------------------------------
Solve for x and y :
--------------------------------------
x = 2y + 3 ----------------------- (1)
10x = 29y - 9 ------------------- (2)

Sub (1) into (2) :

7(2y + 3) = 29y - 9
14y + 21 = 29y - 9
15y = 30
y = 2 ------------------- Sub into (1)
x = 2(2) + 3
x = 7

--------------------------------------
Find the number:
--------------------------------------
Number = 10x + y = 10(2) + 7 = 27

----------------------------------------------------------------------------
Answer: The 2-digit number is 27.
----------------------------------------------------------------------------

4 0

2 years ago